Friday, October 22, 2010

Zzzzzap

So lately I've been taking some time during the day to open up the books and get a bit of a head start for school coming up. A few of the textbooks we require can be used in different levels, which is great. For instance, a couple of our large textbooks cover all kinds of material, so we can use them each time we attend. Some of the smaller, more narrowed down content books we still need to buy, but that's not so bad.

For Year 3 Heavy Duty Mechanics, the subject is Electricity, Carburetor Fuel Systems & Diesel Fuel Systems (Mechanical & Electronic). I'm still not sure why we still cover carburetors, it's pretty antiquated technology. Year 3 is supposedly the hardest of all the 4 levels we have, since it's based mostly on electrical theory. A lot of people struggle with it. I have a pretty good base of understanding of it through work and dabbling a bit in other classes. So far I've been studying the math behind Series, Parallel, and Series-Parallel circuits, which is basically finding Volts, Amps, Ohms and Watts based on diagrams. Back to grade 8 math, really.

There's the formula wheel for what I've been studying. W is for Watts, I is Inductance (Amps), R is Resistance (Ohms) and E is Electromotive Force (Volts). I've been making up a lot of circuits and going through the math to get my head wrapped around it. There's a formula you can use called the Voltage Divider Rule, which allows you to solve for voltage across one or combination of series resistors without first having to solve for the amps. It goes:

(R1 x E) / (R1 + R2 + R3 etc.) = V1

It's fairly simple really. If you're like me and you never do math outside of shopping or cooking and baking, then it's gotta be worked into the brain before I can remember it fully.

So for this particular circuit, looking for the voltage drop across R1, the forumla would read:

(3Ω x 12v) / (3Ω + 5Ω + 4Ω) = V1

Which would then work out to read:

36/12 = 3volts.

So, the voltage drop across R1 is 3 volts. The next formula would read:

(5Ω x 12v) / (3Ω + 5Ω + 4Ω) = V2

60/12 = 5volts.

Voltage drop across R2 is 5 volts. The next formula would read:

(4Ω x 12v) / (3Ω + 5Ω + 4Ω) = V3

48/12 = 4 volts.

I just happened to pick numbers that work out to 12 for the resistance, so it worked out fairly easily. I've been entering all kinds of numbers in my practice to make sure I don't take it too lightly.

So, since in a series circuit, the sum of all resistors equals the total resistance (RT), we can now figure out what the amperage is. In a series circuit, current through each resistor is the same. So the formula goes: